91Ó°ÊÓ

The solution of the given inequality$\left|\frac{4b−2}{3}\right|<12$ is:

Case 1:$\frac{4b−2}{3}$ is non-negative.

$\begin{array}{c}\frac{4b−2}{3}<12\\ \frac{4b−2}{3}×3<12×3\\ 4b−2<36\\ 4b−2+2<36+2\\ 4b<38\\ \frac{4b}{4}<\frac{38}{4}\\ b<9.5\\ b∈\left(−∞,9.5\right)\end{array}$

Case 2:$\frac{4b−2}{3}$ is negative.

$\begin{array}{c}−\left(\frac{4b−2}{3}\right)<12\\ \left(−1\right)\left(−\left(\frac{4b−2}{3}\right)\right)>\left(−1\right)\left(12\right)\\ \frac{4b−2}{3}>−12\\ \frac{4b−2}{3}×3>−12×3\\ 4b−2>−36\\ 4b−2+2>−36+2\\ 4b>−34\\ \frac{4b}{4}>\frac{−34}{4}\\ b>−8.5\\ b∈\left(−8.5,∞\right)\end{array}$

The solution of the inequality$\left|x\right|≤a$ is$xâ‰\yen −a$ and $x≤a$.

That implies the solution of the inequality is the intersection of the solutions of the inequalities$xâ‰\yen −a$ and $x≤a$.

Find the intersection of the solutions of the inequalities$\frac{4b−2}{3}<12$ and$\frac{4b−2}{3}>−12$ to find the solution of the inequality $\left|\frac{4b−2}{3}\right|<12$.

The intersection of the solutions of the inequalities$\frac{4b−2}{3}<12$ and$\frac{4b−2}{3}>−12$ is:

$b∈\left(−8.5,9.5\right)$

Therefore, the solution of the inequality$\left|\frac{4b−2}{3}\right|<12$ is $b∈\left(−8.5,9.5\right)$.

The graph of the solution set which is $b∈\left(−8.5,9.5\right)$is: